Group homomorphism ($\phi (gh)=\phi (g)\phi (h)$)

A group homomorphism is a map $\phi :G\to H$ preserving the group operation: $\phi (gh)=\phi (g)\phi (h)$. It also preserves the identity element and inverse elements. In an additive group, the same requirement reads $\phi (g+h)=\phi (g)+\phi (h)$. A group homomorphism preserves a way of combining inputs, even when several inputs have the same output.

Reduction modulo three from $\mathbb{Z}/6\mathbb{Z}$ to $\mathbb{Z}/3\mathbb{Z}$ is a group homomorphism: reducing a sum agrees with adding the reduced residues. For example, $2+5=1$ modulo six, and both sides reduce to one modulo three; the outputs $2$ and $2$ also add to one modulo three. This map is surjective but not injective, since zero and three both map to zero.

The sign of a permutation gives another example, from the symmetric group $S_3$ to the multiplicative group $\{1,-1\}$. A product of two odd permutations is even, matching $(-1)(-1)=1$. Its group kernel is $A_3$, the alternating group of even permutations. This shows how a group homomorphism can retain a useful feature while forgetting other details.

Let $G$ be a finite group and $\varphi :G\to H$ a surjective group homomorphism. Choose a preimage $g$ of $h$. Since $\varphi (g{)}^{o(g)}=e$, the order of a group element $m=o(h)$ divides $o(g)$; write $o(g)=mr$. Then $g^r$ has order $m$, because $(g^r{)}^s=e$ holds exactly when $mr\mid rs$, or equivalently $m\mid s$. This proves existence of an element of order $m$ in $G$. It need not be a preimage of $h$: for the quotient map ${\mathbb{Z}}_4\to {\mathbb{Z}}_2$, both preimages of the nonidentity target element have order $4$, while the order-$2$ element of the source maps to zero. Thus the assertion concerns element-order existence, not splitting of the quotient map.

A group antihomomorphism between groups reverses multiplication: $f(gh)=f(h)f(g)$. Inversion $g\mapsto g^{-1}$ is an example. An antihomomorphism is a group homomorphism when the multiplication in the target is replaced by its reverse. In a nonabelian target the reversal matters: the group homomorphism and antihomomorphism identities generally differ.

Suppose the cubing map $T:G\to G$ is a group homomorphism. Expanding $(ab{)}^3=a^3{b}^3$ and cancelling gives $(ba{)}^2=a^2{b}^2$. Alsoso $g^2$ commutes with every cube. For $x,y\in T(G)$, therefore, $x^{2}y=y{x}^2$ and $yxyx=x^2{y}^2$. Multiplying the latter identity on the left by $y^{-1}$ yields $xyx=x^{2}y$; cancelling $x$ gives $yx=xy$. Thus the image of a group homomorphism $T(G)$ is an abelian group.

A group homomorphism between abelian groups, conventionally written additively. Its kernel and image are subgroups, and the additive formula implies preservation of zero, negation and every integer multiple. Maps on Mordell-Weil groups induced by isogenies of elliptic curves and by good reduction are examples.

A group homomorphism $f:G\to H$ is surjective when every element of $H$ is $f(g)$ for some $g\in G$.

A covering homomorphism is a surjective homomorphism of topological groups whose underlying map is a covering space. Its discrete kernel is the fiber over the identity.

A homomorphism from $C_n$ is determined by the image of one generator; a surjection $C_n\to C_m$ exists exactly when $m\mid n$.

A group homomorphism from ${\mathbb{Z}}_m$ to the additive cyclic group ${\mathbb{Z}}_n$ is determined by the image $a$ of $1$, subject to $ma\equiv 0(\mathrm{mod}n)$. Conversely every such $a$ gives a well-defined map $k\mapsto ka$. If $d=\gcd (m,n)$, the condition is equivalent to $n/d$ dividing $a$. There are exactly $d$ such residues modulo $n$, proving the formula. More generally, homomorphisms from ${\mathbb{Z}}_m$ to any group correspond to target elements whose order of a group element divides $m$.

The kernel of a group homomorphism is the group version of a kernel. For $\phi :G\to H$, it isIt is a normal subgroup of the domain: for $n$ in the group kernel, $\phi (gn{g}^{-1})=\phi (g)e_H\phi (g{)}^{-1}=e_H$. The group kernel describes inputs whose effect disappears after mapping. For reduction modulo three from $\mathbb{Z}/6\mathbb{Z}$, it is $\{0,3\}$; for the sign of a permutation on $S_3$, it is $A_3$. A group homomorphism is injective exactly when its group kernel contains only the identity element.

For a group homomorphism $\phi :G\to H$, two inputs give the same output exactly when $\phi (h^{-1}g)=e_H$. Consequently the fibre over an output $\phi (h)$ is the left coset $h\ker \phi$. These cosets identify precisely the inputs indistinguishable to $\phi$, which is the mechanism behind the first isomorphism theorem.

The image of a group homomorphism $\varphi :G\to H$ is the subgroup $\{\varphi (g):g\in G\}$ of $H$.